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Euler problems/71 to 80

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[[Category:Programming exercise spoilers]]
 
== [http://projecteuler.net/index.php?section=view&id=71 Problem 71] ==
== [http://projecteuler.net/index.php?section=view&id=71 Problem 71] ==
Listing reduced proper fractions in ascending order of size.
Listing reduced proper fractions in ascending order of size.
Line 5: Line 4:
Solution:
Solution:
<haskell>
<haskell>
-
import Data.Ratio (Ratio, (%), numerator)
+
-- http://mathworld.wolfram.com/FareySequence.html
-
 
+
import Data.Ratio ((%), numerator,denominator)
-
fractions :: [Ratio Integer]
+
fareySeq a b
-
fractions = [f | d <- [1..1000000], let n = (d * 3) `div` 7, let f = n%d, f /= 3%7]
+
|da2<=10^6=fareySeq a1 b
-
 
+
|otherwise=na
-
problem_71 :: Integer
+
where
-
problem_71 = numerator $ maximum $ fractions
+
na=numerator a
 +
nb=numerator b
 +
da=denominator a
 +
db=denominator b
 +
a1=(na+nb)%(da+db)
 +
da2=denominator a1
 +
problem_71=fareySeq (0%1) (3%7)
</haskell>
</haskell>
Line 20: Line 25:
Using the [http://mathworld.wolfram.com/FareySequence.html Farey Sequence] method, the solution is the sum of phi (n) from 1 to 1000000.
Using the [http://mathworld.wolfram.com/FareySequence.html Farey Sequence] method, the solution is the sum of phi (n) from 1 to 1000000.
-
 
-
See problem 69 for phi function
 
-
 
<haskell>
<haskell>
-
problem_72 = sum [phi x|x <- [1..1000000]]
+
groups=1000
 +
eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors)
 +
where factors = fstfac n
 +
fstfac x = [(head a ,length a)|a<-group$primeFactors x]
 +
p72 n= sum [eulerTotient x|x <- [groups*n+1..groups*(n+1)]]
 +
problem_72 = sum [p72 x|x <- [0..999]]
</haskell>
</haskell>
Line 31: Line 38:
Solution:
Solution:
-
<haskell>
 
-
import Data.Ratio (Ratio, (%), numerator, denominator)
 
-
median :: Ratio Int -> Ratio Int -> Ratio Int
+
If you haven't done so already, read about Farey sequences in Wikipedia
-
median a b = ((numerator a) + (numerator b)) % ((denominator a) + (denominator b))
+
http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about
 +
mediants. Then divide and conquer. The number of Farey ratios between
 +
(a, b) is 1 + the number between (a, mediant a b) + the number between
 +
(mediant a b, b). Henrylaxen 2008-03-04
-
count :: Ratio Int -> Ratio Int -> Int
+
<haskell>
-
count a b
+
import Data.Ratio
-
| d > 10000 = 1
+
-
| otherwise = count a m + count m b
+
-
where
+
-
m = median a b
+
-
d = denominator m
+
-
problem_73 :: Int
+
mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a
-
problem_73 = (count (1%3) (1%2)) - 1
+
mediant f1 f2 = (numerator f1 + numerator f2) %
 +
(denominator f1 + denominator f2)
 +
fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t
 +
fareyCount n (a,b) =
 +
let c = mediant a b
 +
in if (denominator c > n) then 0 else
 +
1 + (fareyCount n (a,c)) + (fareyCount n (c,b))
 +
 +
problem_73 :: Integer
 +
problem_73 = fareyCount 10000 (1%3,1%2)
</haskell>
</haskell>
 +
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
Line 54: Line 67:
Solution:
Solution:
<haskell>
<haskell>
-
import Data.Array (Array, array, (!), elems)
+
import Data.List
-
import Data.Char (ord)
+
explode 0 = []
-
import Data.List (foldl1')
+
explode n = n `mod` 10 : explode (n `quot` 10)
-
import Prelude hiding (cycle)
+
-
 
+
chain 2 = 1
-
fact :: Integer -> Integer
+
chain 1 = 1
-
fact 0 = 1
+
chain 145 = 1
-
fact n = foldl1' (*) [1..n]
+
chain 40585 = 1
-
 
+
chain 169 = 3
-
factorDigits :: Array Integer Integer
+
chain 363601 = 3
-
factorDigits = array (0,2177281) [(x,n)|x <- [0..2177281], let n = sum $ map (\y -> fact (toInteger $ ord y - 48)) $ show x]
+
chain 1454 = 3
-
 
+
chain 871 = 2
-
cycle :: Integer -> Integer
+
chain 45361 = 2
-
cycle 145 = 1
+
chain 872 = 2
-
cycle 169 = 3
+
chain 45362 = 2
-
cycle 363601 = 3
+
chain x = 1 + chain (sumFactDigits x)
-
cycle 1454 = 3
+
makeIncreas 1 minnum = [[a]|a<-[minnum..9]]
-
cycle 871 = 2
+
makeIncreas digits minnum = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a]
-
cycle 45361 = 2
+
p74=
-
cycle 872 = 2
+
sum[div p6 $countNum a|
-
cycle 45362 = 2
+
a<-tail$makeIncreas 6 1,
-
cycle _ = 0
+
let k=digitToN a,
-
 
+
chain k==60
-
isChainLength :: Integer -> Integer -> Bool
+
]
-
isChainLength len n
+
-
| len < 0 = False
+
-
| t = isChainLength (len-1) n'
+
-
| otherwise = (len - c) == 0
+
where
where
-
c = cycle n
+
p6=facts!! 6
-
t = c == 0
+
sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode
-
n' = factorDigits ! n
+
factorial n = if n == 0 then 1 else n * factorial (n - 1)
-
 
+
digitToN = foldl' (\a b -> 10*a + b) 0 .dropWhile (==0)
-
-- | strict version of the maximum function
+
facts = scanl (*) 1 [1..9]
-
maximum' :: (Ord a) => [a] -> a
+
countNum xs=ys
-
maximum' [] = undefined
+
where
-
maximum' [x] = x
+
ys=product$map (factorial.length)$group xs
-
maximum' (a:b:xs) = let m = max a b in m `seq` maximum' (m : xs)
+
problem_74= length[k|k<-[1..9999],chain k==60]+p74
-
 
+
test = print $ [a|a<-tail$makeIncreas 6 0,let k=digitToN a,chain k==60]
-
problem_74 :: Int
+
-
problem_74 = length $ filter (\(_,b) -> isChainLength 59 b) $ zip ([0..] :: [Integer]) $ take 1000000 $ elems factorDigits
+
</haskell>
</haskell>
-
 
== [http://projecteuler.net/index.php?section=view&id=75 Problem 75] ==
== [http://projecteuler.net/index.php?section=view&id=75 Problem 75] ==
Find the number of different lengths of wire can that can form a right angle triangle in only one way.
Find the number of different lengths of wire can that can form a right angle triangle in only one way.
Solution:
Solution:
-
This is only slightly harder than [[Euler problems/31 to 40#39|problem 39]]. The search condition is simpler but the search space is larger.
 
<haskell>
<haskell>
-
problem_75 = length . filter ((== 1) . length) $ group perims
+
import Data.Array
-
where perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]]
+
 
-
pTriples = [p |
+
triangs :: [Int]
-
n <- [1..1000],
+
triangs = [p | n <- [2..1000],
-
m <- [n+1..1000],
+
m <- [1..n-1],
-
even n || even m,
+
gcd m n == 1,
-
gcd n m == 1,
+
odd (m+n),
-
let a = m^2 - n^2,
+
let p = 2 * (n^2 + m*n),
-
let b = 2*m*n,
+
p <= 2*10^6]
-
let c = m^2 + n^2,
+
 
-
let p = a + b + c,
+
problem_75 :: Int
-
p <= 10^6]
+
problem_75 = length
 +
$ filter (\(_, c) -> c == 1)
 +
$ assocs
 +
$ (\ns -> accumArray (+) 0 (1, 2*10^6) [(n, 1) | n <- ns, inRange (1, 2*10^6) n])
 +
$ concatMap (\n -> takeWhile (<=2*10^6) [n,2*n..]) triangs
</haskell>
</haskell>
Line 122: Line 131:
Solution:
Solution:
-
Calculated using Euler's pentagonal formula and a list for memoization.
+
Here is a simpler solution: For each n, we create the list of the number of partitions of n
 +
whose lowest number is i, for i=1..n. We build up the list of these lists for n=0..100.
<haskell>
<haskell>
-
partitions = 1 : [sum [s * partitions !! p| (s,p) <- zip signs $ parts n]| n <- [1..]]
+
build x = (map sum (zipWith drop [0..] x) ++ [1]) : x
-
where
+
problem_76 = (sum $ head $ iterate build [] !! 100) - 1
-
signs = cycle [1,1,(-1),(-1)]
+
-
suite = map penta $ concat [[n,(-n)]|n <- [1..]]
+
-
penta n = n*(3*n - 1) `div` 2
+
-
parts n = takeWhile (>= 0) [n-x| x <- suite]
+
-
 
+
-
problem_76 = partitions !! 100 - 1
+
</haskell>
</haskell>
Line 141: Line 145:
Brute force but still finds the solution in less than one second.
Brute force but still finds the solution in less than one second.
<haskell>
<haskell>
-
combinations acc 0 _ = [acc]
+
counter = foldl (\without p ->
-
combinations acc _ [] = []
+
let (poor,rich) = splitAt p without
-
combinations acc value prim@(x:xs) = combinations (acc ++ [x]) value' prim' ++ combinations acc value xs
+
with = poor ++
-
where
+
zipWith (+) with rich
-
value' = value - x
+
in with
-
prim' = dropWhile (>value') prim
+
) (1 : repeat 0)
-
 
+
-
problem_77 :: Integer
+
problem_77 =
-
problem_77 = head $ filter f [1..]
+
find ((>5000) . (ways !!)) $ [1..]
where
where
-
f n = (length $ combinations [] n $ takeWhile (<n) primes) > 5000
+
ways = counter $ take 100 primes
</haskell>
</haskell>
Line 158: Line 162:
Solution:
Solution:
-
 
-
Same as problem 76 but using array instead of lists to speedup things.
 
<haskell>
<haskell>
import Data.Array
import Data.Array
partitions :: Array Int Integer
partitions :: Array Int Integer
-
partitions = array (0,1000000) $ (0,1) : [(n,sum [s * partitions ! p| (s,p) <- zip signs $ parts n])| n <- [1..1000000]]
+
partitions =
 +
array (0,1000000) $
 +
(0,1) :
 +
[(n,sum [s * partitions ! p|
 +
(s,p) <- zip signs $ parts n])|
 +
n <- [1..1000000]]
where
where
signs = cycle [1,1,(-1),(-1)]
signs = cycle [1,1,(-1),(-1)]
Line 172: Line 179:
problem_78 :: Int
problem_78 :: Int
-
problem_78 = head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..]
+
problem_78 =
 +
head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..]
</haskell>
</haskell>
Line 179: Line 187:
Solution:
Solution:
-
 
-
A bit ugly but works fine
 
<haskell>
<haskell>
-
import Data.List
+
import Data.Char (digitToInt, intToDigit)
-
 
+
import Data.Graph (buildG, topSort)
-
problem_79 :: String -> String
+
import Data.List (intersect)
-
problem_79 file = map fst $ sortBy (\(_,a) (_,b) -> compare (length b) (length a)) $ zip digs order
+
 +
p79 file=
 +
(+0)$read . intersect graphWalk $ usedDigits
where
where
-
nums = lines file
+
usedDigits = intersect "0123456789" $ file
-
digs = map head $ group $ sort $ filter (\c -> c >= '0' && c <= '9') file
+
edges = concatMap (edgePair . map digitToInt) . words $ file
-
prec = concatMap (\(x:y:z:_) -> [[x,y],[y,z],[x,z]]) nums
+
graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges
-
order = map (\n -> map head $ group $ sort $ map (\(_:x:_) -> x) $ filter (\(x:_) -> x == n) prec) digs
+
edgePair [x, y, z] = [(x, y), (y, z)]
 +
edgePair _ = undefined
 +
 +
problem_79 = do
 +
f<-readFile "keylog.txt"
 +
print $p79 f
</haskell>
</haskell>
Line 196: Line 209:
Calculating the digital sum of the decimal digits of irrational square roots.
Calculating the digital sum of the decimal digits of irrational square roots.
-
Solution:
+
This solution uses binary search to find the square root of a large Integer:
<haskell>
<haskell>
-
module Main where
+
import Data.Char (digitToInt)
-
import Data.List ((\\))
+
intSqrt :: Integer -> Integer
-
 
+
intSqrt n = bsearch 1 n
-
hundreds :: Integer -> [Integer]
+
-
hundreds n = hundreds' [] n
+
where
where
-
hundreds' acc 0 = acc
+
bsearch l u = let m = (l+u) `div` 2
-
hundreds' acc n = hundreds' (m : acc) d
+
m2 = m^2
-
where
+
in if u <= l
-
(d,m) = divMod n 100
+
then m
 +
else if m2 < n
 +
then bsearch (m+1) u
 +
else bsearch l m
-
squareDigs :: Integer -> [Integer]
+
problem_80 :: Int
-
squareDigs n = p : squareDigs' p r xs
+
problem_80 = sum [f r | a <- [1..100],
 +
let x = a * e,
 +
let r = intSqrt x,
 +
r*r /= x]
where
where
-
(x:xs) = hundreds n ++ repeat 0
+
e = 10^202
-
p = floor $ sqrt $ fromInteger x
+
f = sum . take 100 . map digitToInt . show
-
r = x - (p^2)
+
-
 
+
-
squareDigs' :: Integer -> Integer -> [Integer] -> [Integer]
+
-
squareDigs' p r (x:xs) = x' : squareDigs' (p*10 + x') r' xs
+
-
where
+
-
n = 100*r + x
+
-
(x',r') = last $ takeWhile (\(_,a) -> a >= 0) $ scanl (\(_,b) (a',b') -> (a',b-b')) (0,n) rs
+
-
rs = [y|y <- zip [1..] [(20*p+1),(20*p+3)..]]
+
-
+
-
sumDigits n = sum $ take 100 $ squareDigs n
+
-
 
+
-
problem_80 :: Integer
+
-
problem_80 = sum $ map sumDigits [x|x <- [1..100] \\ [n^2|n<-[1..10]]]
+
</haskell>
</haskell>
-
 
-
[[Category:Tutorials]]
 
-
[[Category:Code]]
 

Current revision

Contents

1 Problem 71

Listing reduced proper fractions in ascending order of size.

Solution: 

-- http://mathworld.wolfram.com/FareySequence.html 
import Data.Ratio ((%), numerator,denominator)
fareySeq a b
    |da2<=10^6=fareySeq a1 b
    |otherwise=na
    where
    na=numerator a
    nb=numerator b
    da=denominator a
    db=denominator b
    a1=(na+nb)%(da+db)
    da2=denominator a1
problem_71=fareySeq (0%1) (3%7)

2 Problem 72

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

Solution:

Using the Farey Sequence method, the solution is the sum of phi (n) from 1 to 1000000. 

groups=1000
eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors)
    where factors = fstfac n
fstfac x = [(head a ,length a)|a<-group$primeFactors x] 
p72 n= sum [eulerTotient x|x <- [groups*n+1..groups*(n+1)]]
problem_72 = sum [p72 x|x <- [0..999]]

3 Problem 73

How many fractions lie between 1/3 and 1/2 in a sorted set of reduced proper fractions?

Solution:

If you haven't done so already, read about Farey sequences in Wikipedia http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about mediants. Then divide and conquer. The number of Farey ratios between (a, b) is 1 + the number between (a, mediant a b) + the number between (mediant a b, b). Henrylaxen 2008-03-04 

import Data.Ratio
 
mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a
mediant f1 f2 = (numerator f1 + numerator f2) % 
                (denominator f1 + denominator f2)
fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t
fareyCount n (a,b) =
  let c = mediant a b
  in  if (denominator c > n) then 0 else 
         1 + (fareyCount n (a,c)) + (fareyCount n (c,b))
 
problem_73 :: Integer
problem_73 =  fareyCount 10000   (1%3,1%2)


4 Problem 74

Determine the number of factorial chains that contain exactly sixty non-repeating terms.

Solution: 

import Data.List
explode 0 = []
explode n = n `mod` 10 : explode (n `quot` 10)
 
chain 2    = 1
chain 1    = 1
chain 145    = 1
chain 40585    = 1
chain 169    = 3
chain 363601 = 3
chain 1454   = 3
chain 871    = 2
chain 45361  = 2
chain 872    = 2
chain 45362  = 2
chain x = 1 + chain (sumFactDigits x)
makeIncreas 1 minnum  = [[a]|a<-[minnum..9]]
makeIncreas digits minnum  = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a]
p74=
    sum[div p6 $countNum a|
    a<-tail$makeIncreas  6 1,
    let k=digitToN a,
    chain k==60
    ]
    where
    p6=facts!! 6
sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode
factorial n = if n == 0 then 1 else n * factorial (n - 1)
digitToN = foldl' (\a b -> 10*a + b) 0 .dropWhile (==0)
facts = scanl (*) 1 [1..9]
countNum xs=ys
    where
    ys=product$map (factorial.length)$group xs 
problem_74= length[k|k<-[1..9999],chain k==60]+p74
test = print $ [a|a<-tail$makeIncreas 6 0,let k=digitToN a,chain k==60]

5 Problem 75

Find the number of different lengths of wire can that can form a right angle triangle in only one way.

Solution: 

import Data.Array
 
triangs :: [Int]
triangs = [p | n <- [2..1000],
               m <- [1..n-1],
               gcd m n == 1,
               odd (m+n),
               let p = 2 * (n^2 + m*n),
               p <= 2*10^6]
 
problem_75 :: Int
problem_75 = length
       $ filter (\(_, c) -> c == 1)
       $ assocs
       $ (\ns -> accumArray (+) 0 (1, 2*10^6) [(n, 1) | n <- ns, inRange (1, 2*10^6) n])
       $ concatMap (\n -> takeWhile (<=2*10^6) [n,2*n..]) triangs

6 Problem 76

How many different ways can one hundred be written as a sum of at least two positive integers?

Solution:

Here is a simpler solution: For each n, we create the list of the number of partitions of n whose lowest number is i, for i=1..n. We build up the list of these lists for n=0..100. 

build x = (map sum (zipWith drop [0..] x) ++ [1]) : x
problem_76 = (sum $ head $ iterate build [] !! 100) - 1

7 Problem 77

What is the first value which can be written as the sum of primes in over five thousand different ways?

Solution:

Brute force but still finds the solution in less than one second. 

counter = foldl (\without p ->
                     let (poor,rich) = splitAt p without
                         with = poor ++ 
                                zipWith (+) with rich
                     in with
                ) (1 : repeat 0)
 
problem_77 =  
    find ((>5000) . (ways !!)) $ [1..]
    where
    ways = counter $ take 100 primes

8 Problem 78

Investigating the number of ways in which coins can be separated into piles.

Solution: 

import Data.Array
 
partitions :: Array Int Integer
partitions = 
    array (0,1000000) $ 
    (0,1) : 
    [(n,sum [s * partitions ! p|
    (s,p) <- zip signs $ parts n])|
    n <- [1..1000000]]
    where
        signs = cycle [1,1,(-1),(-1)]
        suite = map penta $ concat [[n,(-n)]|n <- [1..]]
        penta n = n*(3*n - 1) `div` 2
        parts n = takeWhile (>= 0) [n-x| x <- suite]
 
problem_78 :: Int
problem_78 = 
    head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..]

9 Problem 79

By analysing a user's login attempts, can you determine the secret numeric passcode?

Solution: 

import Data.Char (digitToInt, intToDigit)
import Data.Graph (buildG, topSort)
import Data.List (intersect)
 
p79 file= 
    (+0)$read . intersect graphWalk $ usedDigits
    where
    usedDigits = intersect "0123456789" $ file
    edges = concatMap (edgePair . map digitToInt) . words $ file
    graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges
    edgePair [x, y, z] = [(x, y), (y, z)]
    edgePair _         = undefined
 
problem_79 = do
    f<-readFile  "keylog.txt"
    print $p79 f

10 Problem 80

Calculating the digital sum of the decimal digits of irrational square roots.

This solution uses binary search to find the square root of a large Integer: 

import Data.Char (digitToInt)
 
intSqrt :: Integer -> Integer
intSqrt n = bsearch 1 n
    where
      bsearch l u = let m = (l+u) `div` 2
                        m2 = m^2
                    in if u <= l
                       then m
                       else if m2 < n
                            then bsearch (m+1) u
                            else bsearch l m
 
problem_80 :: Int
problem_80 = sum [f r | a <- [1..100],
                        let x = a * e,
                        let r = intSqrt x,
                        r*r /= x]
    where
      e = 10^202
      f = sum . take 100 . map digitToInt . show
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/71_to_80"